Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{-r - 1}{-3r^2 - 24r - 21} \times \dfrac{r^2 + 6r - 7}{r - 5} $
Solution: First factor out any common factors. $q = \dfrac{-(r + 1)}{-3(r^2 + 8r + 7)} \times \dfrac{r^2 + 6r - 7}{r - 5} $ Then factor the quadratic expressions. $q = \dfrac {-(r + 1)} {-3(r + 7)(r + 1)} \times \dfrac {(r + 7)(r - 1)} {r - 5} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {-(r + 1) \times (r + 7)(r - 1) } { -3(r + 7)(r + 1) \times (r - 5)} $ $q = \dfrac {-(r + 7)(r - 1)(r + 1)} {-3(r + 7)(r + 1)(r - 5)} $ Notice that $(r + 7)$ and $(r + 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {-\cancel{(r + 7)}(r - 1)(r + 1)} {-3\cancel{(r + 7)}(r + 1)(r - 5)} $ We are dividing by $r + 7$ , so $r + 7 \neq 0$ Therefore, $r \neq -7$ $q = \dfrac {-\cancel{(r + 7)}(r - 1)\cancel{(r + 1)}} {-3\cancel{(r + 7)}\cancel{(r + 1)}(r - 5)} $ We are dividing by $r + 1$ , so $r + 1 \neq 0$ Therefore, $r \neq -1$ $q = \dfrac {-(r - 1)} {-3(r - 5)} $ $ q = \dfrac{r - 1}{3(r - 5)}; r \neq -7; r \neq -1 $